SOLUTION: The half life of cobalt -60 is 5.27 years. Starting with a sample of 150 mg, after how many years is 20 mg left?

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Question 1191731: The half life of cobalt -60 is 5.27 years. Starting with a sample of 150 mg, after how many years is 20 mg left?
Found 4 solutions by josgarithmetic, Theo, greenestamps, ikleyn:
Answer by josgarithmetic(39613) About Me  (Show Source):
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Q=150%2A%280.5%29%5E%28t%2F5.27%29

- - - -

log%28%28Q%29%29=log%28%28150%280.5%29%5E%28t%2F5.27%29%29%29

log%28%28Q%29%29=log%28%28150%29%29%2Blog%28%280.5%5E%28t%2F5.27%29%29%29

log%28%28Q%29%29=log%28%28150%29%29%2B%28t%2F5.27%29log%28%280.5%29%29

log%28%28Q%29%29-log%28%28150%29%29=%28t%2F5.27%29log%28%280.5%29%29

5.27%28log%28%28Q%29%29-log%28%28150%29%29%29%2Flog%28%280.5%29%29=t

Description's question is find t when Q=20.

t=5.27%28log%28%2820%29%29-log%28%28150%29%29%29%2Flog%28%280.5%29%29

Answer by Theo(13342) About Me  (Show Source):
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one of the formula you can use is:
f = p * (1 + r) ^ n
when f = 1 and p = 2 and n = 5.27, you get:
1 = 2 * (1 + r) ^ 5.27
divide both sides of this equation by 2 to get:
.5 = (1 + r) ^ 5.27
take the 5.27th root of both sides of this equation to get:
.5 ^ (1/5.27) = (1 + r)
solve for (1 + r) to get:
(1 + r) = .8767556206.
that's your growth factor per year.

when p = 150 and f = 20, the formula becomes:
20 = 150 * .8767556206 ^ n
divide both sides of this equation by 150 to get:
20/150 = .8767556206 ^ n
take the log of both sides of this equation to get:
log(20/150) = n * log(.8767556206).
divide both sides by log(.8767556206) to get:
log(20/150) / log(.8767556206) = n
solve for n to get:
n = 15.31931344.

.8767556206 and 15.31931344 are rounded to the number of digits that the calculator can display.
i stored these number into memory, which take the answer out to more decimal digits.
this makes it more accurate.
i used the stored number to confirm the answer is correct, even though i am showing you the displayed numbers

2 * .8667556206 ^ 5.27 = 1
1 is 1/2 of 2, so that's your half life.
the growth factor per year is .8667556206.
2 will shrink to 1 in 5.27 years at a yearly growth rate of .8667556206.

150 * .8667556206 ^ 15.31931344 = 20.
150 will shrink to 20 in 15.31931344 years at a yearly growth rate of .8667556206.

note that (1 + r) = .8667556206.
solve for r to get:
r = .8667556206 - 1 = -.1232443794.
that's your annual growth rate.
the life of cobalt -60 is being reduced by approximately 12.32% each year.

the equation can be graphed.
it looks like this.



in the graph, you can see that the remaining life of cobalt -60 if 150 years at the beginning, then 75 after 5.27 years, then 20 after 15.3931344 years.










Answer by greenestamps(13195) About Me  (Show Source):
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Both responses you have received show the same right answer; but the calculations performed in each response are far more convoluted than what is necessary.

You have 150mg of a substance decaying to 20mg in an unknown number of half-lives. The equation is simple:

150%28%281%2F2%29%5En%29=20
150%2F20+=+2%5En

The variable is in an exponent, so use logarithms and use a calculator.

n%2Alog%282%29=log%287.5%29
n+=+log%287.5%29%2Flog%282%29

That number of half-lives, to several decimal places, is 2.90689. Multiply that by the 5.27 years for one half life and you get the answer of approximately 15.32 years.

------------------------------------------------------------------------

It should be noted that radioactive decay is a statistical process; after one half life the amount remaining is APPROXIMATELY half of the original. So keeping 6 or 7 decimal places in the answer to a problem like this is unreasonable.

Answer by ikleyn(52754) About Me  (Show Source):
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.
The half life of cobalt -60 is 5.27 years. Starting with a sample of 150 mg, after how many years is 20 mg left?
~~~~~~~~~~~~~~~~ 

In terms of the half-life, the general formula for radioactive decay of cobalt-60 is

    M(t) = M%280%29.%281%2F2%29%5E%28t%2F5.27%29

where M(t) is the current mass of the cobalt-60; M(0) is the initial mass,



Since 20 mg of the cobalt-60 remained, you have this equation

    20 = 150%2A%281%2F2%29%5E%28t%2F5.27%29,  which reduces to   20%2F150 = %281%2F2%29%5E%28t%2F5.27%29,

or

    0.133333 = %281%2F2%29%5E%28t%2F5.27%29.
    


To solve it, take logarithm base 10 from both sides. You get an equation 

    log%2810%2C%280.133333%29%29 = %28t%2F5.27%29%2Alog%2810%2C%280.5%29%29 .


Therefore,

     t = 5.27%2A%28log%2810%2C%280.133333%29%29%2Flog%2810%2C%280.5%29%29%29 = 15.32 years  (rounded)


ANSWER.  It will happen in 15.32 years.

Solved.


Regarding precision of the solution,  if the half-life is given with two decimals,
there is no sense to write or calculate the final sough time with the greater precision.

The greater precision is  a  FICTION  in this case.

------------------

On radioactive decay,  see the lesson
    - Radioactive decay problems
in this site.

You will find many similar  (and different)  solved problems there.


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    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Logarithms".


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